SQL count distinct与NULL

$sh@ORCL> select * from tt2;    TT2_ID TT2_NAME   TT2_DATE---------- ---------- ---------         1 1          23-DEC-13         2            23-DEC-13         3            23-DEC-13         4 1          23-DEC-13$sh@ORCL> SELECT count(tt2_name) FROM tt2;COUNT(TT2_NAME)---------------              2$sh@ORCL> select distinct tt2_name from tt2;TT2_NAME----------1$sh@ORCL> select sum(tt2_name) from tt2;SUM(TT2_NAME)-------------            2 $sh@ORCL> select avg(tt2_name) from tt2;AVG(TT2_NAME)-------------            1

结论：

null不参与count运算，同样适用于sum, avg。

distinct时，所有null作为一个值。